Consider the field of rational numbers $\mathbb{Q}$, and a finite number of complex numbers $c_1, \ldots, c_\ell$.
My question is: what is the automorphism group of $\mathbb{Q}(c_1,\ldots,c_\ell)$ ? Is there a fairly good general expression for this group, or is this question too hard to crack ?
This is way too hard, even if all the $c_i$ are algebraic.
Every finite extension of $\mathbb{Q}$ can be realized as $\mathbb{Q}(c)$ for some complex number $c$; this is the Primitive Element Theorem. In particular, every finite Galois extension of $\mathbb{Q}$ is of this form.
The Inverse Galois Problem asks whether every finite group occurs as the Galois group of some finite extension of $\mathbb{Q}$; hence, as $\mathrm{Aut}(\mathbb{Q}(c))$ for suitably chosen $c\in\mathbb{C}$. This is still open, but we know that at least every finite solvable group can be so realized (this is a theorem of Shafarevich). It is also known that all sporadic simple groups (except perhaps $M_{22}$) is realizable this way, and in addition many other simple groups.
So the class of groups you are contemplating includes at least all finite solvable groups, potentially all finite groups. If we also allow transcendental numbers, since the transcendental degree of $\mathbb{C}$ over $\mathbb{Q}$ is uncountable, you also have automorphism groups of purely transcendental extensions of any extension of finitie degree (this includes $\mathrm{PGL}(2,\mathbb{Q})$ for $\mathbb{Q}(\pi)$, for example), this only complicates matters further.